Core 4 Workshop Download - Elite · PDF fileELITE Tuition, 1 Castle Road, Northolt,...

ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD Tel: 0800 612 9545 Email: [email protected] Web: www.elitetuition.com

Core 3 / Core 4 Workshops

A* Booster Course Dear Student, ELITE Tuition is running a series of two day, EdExcel Mathematics C3 / C4 workshops, designed to ensure students know all of the examination tricks necessary to secure the A*. Students are taught in small groups by an EdExcel Examiner, paying specific attention to exam technique and how to answer the A* grade questions.

The cost of the workshop, for both days including all workshop material is £250.00. However, students from West London schools are eligible for a 40% discount, making the cost £150.00. Each day will look, into detail, 60 examination questions from a bank of EdExcel questions not commonly available to students.

The two-day workshops will be running in early June 2011. The structure of the days will be as follows: Day 1: Core 3 11th June Day 2: Core 4 18th June 11:00am Introduction 11:00am Introduction 11:30am C3 Workshop Part 1 11:30am C4 Workshop Part 1 1:30pm Analysis of Trick Questions 1:30pm Analysis of Trick Questions 2:00pm Lunch 2:00pm Lunch 3:00pm C3 Workshop Part 2 3:00pm C4 Workshop Part 2 5:00pm Top Tips for C3 Success 5:00pm Top Tips for C4 Success 5:30pm Finish 5:30pm Finish If you are interested in attending, please fill in the section below and enclose a cheque of £150.00 made payable to ELITE Tuition, and send these to ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD.

For further information, please call 0800 612 9545 or visit www.elitetuition.com.

Best wishes, Rachael Gill (School Liaisons Officer)

All details must be completed below.

Surname of Student: First Name(s):

Date of Birth: / / Email:

Home Address:

Post Code:

Home Telephone Number: Student Mobile:

Name of School / College:

I confirm that I will be attending the ELITE Tuition EdExcel Core 3 / Core 4 Revision Workshop “A* Booster Course” on the Saturdays 11th and 18th June at the venue selected: Merchant Taylors School. I enclose a cheque of sum £150.00 made payable to ELITE Tuition. Signed: Print Name:


C4 Binomial Expansion Problem Question:

The function f (x) is defined as f x( ) ≡ 6xx2 − 4x + 3

, x < 1 .

a) Express f (x) as partial fractions. (3 marks)

b) Show that for small values of x,

f x( ) ≈ 2x + 83x2 + 26

9x3

(5 marks)

Answer: a) Writing as partial fractions (assuming not top heavy):

f x( ) ≡ 6xx2 − 4x + 3 (Factoring the denominator)

x2 − 4x + 3 ≡ x −1( ) x − 3( )

⇒ f x( ) ≡ 6xx −1( ) x − 3( ) ≡

Ax −1( ) +

Bx − 3( )

Multiplying through: 6x ≡ A x − 3( ) + B x −1( ) Select values of x to eliminate A and B. Let x = 1 6 1( ) = A 1− 3( ) + B 1−1( ) 6 = −2A + B 0( ) ∴A = −3 Let x = 3 6 3( ) = A 3− 3( ) + B 3−1( ) 18 = A 0( ) + 2B ∴B = 9

⇒ f x( ) ≡ 9x − 3( ) −

3x −1( )

Please turn over…


b) Re-writing our expression for f (x):

f x( ) ≡ 9 x − 3( )−1 − 3 x −1( )−1 We cannot perform a binomial expansion here unless it is in the form 1+ ax( )n . So if we re-write it a little more: f x( ) ≡ 9 −3+ x( )−1 − 3 −1+ x( )−1 Factoring to put into the form of 1+ ax( )n :

f x( ) ≡ 9 × −3−1 1− x3

⎛⎝⎜

⎞⎠⎟−1

− 3× −1−1 1− x1

⎛⎝⎜

⎞⎠⎟−1

f x( ) ≡ 9 × −131− x

3⎛⎝⎜

⎞⎠⎟−1

− 3× −111− x

1⎛⎝⎜

⎞⎠⎟−1

f x( ) ≡ −3 1− x3

⎛⎝⎜

⎞⎠⎟−1

+ 3 1− x1

⎛⎝⎜

⎞⎠⎟−1

∴ f x( ) ≡ 3 1− x( )−1 − 3 1− x3

⎛⎝⎜

⎞⎠⎟−1

Expanding the first term: Use 1− x( )−1 = 1+ x + x2 + x3 + x4 + x5 + ... 3 1− x( )−1 = 3 1+ x + x2 + x3 + ...⎡⎣ ⎤⎦ 3 1− x( )−1 = 3+ 3x + 3x2 + 3x3 + ... Expanding the second term:

−3 1− x3

⎛⎝⎜

⎞⎠⎟−1

= −3 1+ x3+x2

9+x3

27+ ...

⎡

⎣⎢

⎤

⎦⎥

−3 1− x3

⎛⎝⎜

⎞⎠⎟−1

= −3+ −x − x2

3−x3

9+ ...

Adding the two results:

f x( ) ≡ 3+ 3x + 3x2 + 3x3 + ...− 3+ −x − x2

3−x3

9+ ... = 2x + 8

3x2 + 26

9x3 + ...

Yet when x is small:

∴ f x( ) ≈ 2x + 83x2 + 26

9x3

Found this useful? Attend our A* Booster C3 / C4 Workshop.


C4 Parametric Integration Problem Question:

An ellipse is defined be the following parametric equations x = 4 cosθ , y = 2sinθ , 0 ≤θ ≤ 2π .

which meets the positive coordinate axes at the points A and B c) Find the values of the parameter θ at the points A and B.

(2 marks)

d) Show that the area of the ellipse is 8π. (8 marks)

Answer: a) It is always useful to draw a diagram when solving problems.

Hence, A is found when x = 0 and B is found when y = 0. Thus: When x = 0 When y = 0 0 = 4 cosθ 0 = 2sinθ

⇒θ =π2 ⇒θ = 0

b) Using the equation for parametric equations:

y dxa

b

∫ = y dxdθ

dθa '

b '

∫

y = 2sinθ & dxdθ

= −4sinθ ⇒ Area = 2sinθ × −4sinθ dθπ

2

0

∫

Please turn over…


⇒ Area = −8sin2θ dθπ

2

0

∫ (Remove – sign and flip limits)

⇒ Area = 8sin2θ dθ0

π2

∫ (Use sin2θ =121− cos2θ( ) )

Area = 8 ×12

1− cos2θ( ) dθ0

π2

∫

Area = 4 1− cos2θ dθ0

π2

∫ (Integrate)

Area = 4 θ −12sin2θ⎡

⎣⎢⎤⎦⎥0

π2

Area = 4 π2−12sin2 π

2⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥− 0 − 1

2sin2 0( )⎡

⎣⎢⎤⎦⎥

⎧⎨⎩

⎫⎬⎭

Area = 4 π2− 0⎡

⎣⎢⎤⎦⎥− 0 − 0[ ]⎧

⎨⎩

⎫⎬⎭

Area = 2π Since the shaded area is a quarter of the area of the ellipse, the total area of the ellipse would be 8π. ∴Total Area = 8π



C4 Differential Equation Problem Question:

Given that y = 0 when x = 0, solve the differential equation: dydx

= ex+ y cos x

(10 marks)

Answer:

First one needs to separate the variables: dydx

= ex+ y cos x (Use ex+ y = exey )

dydx

= exey cos x (Separate the variables)

e− y dy = ex cos x dx (Integrate both sides) e− y dy = ex cos x dx∫∫

Integrating the left hand side first: −e− y = ex cos x dx∫ (Ignoring the constant from LHS) Note: If you ignore the integrating constant from the left hand side, you MUST remember to include it when integrating the right hand side. Integrating the right hand side by parts:

Let: u = ex dvdx

= cos x

dudx

= ex v = sin x

Use I = uv − v dudx

dx∫ from the C4 Formula Bookmark:

−e− y = ex sin x − ex sin x dx∫

Please turn over…


Integrate ex sin x dx∫ by parts again:

Let: u = ex dvdx

= sin x

dudx

= ex v = − cos x

Substituting back in: −e− y = ex sin x − −ex cos x − −ex cos x dx∫⎡

⎣⎤⎦

Simplifying the negative signs. −e− y = ex sin x − −ex cos x + ex cos x dx∫⎡

⎣⎤⎦

−e− y = ex sin x + ex cos x − ex cos x dx∫

The trick behind this question is to notice that ex cos x dx∫ is −e− y :

Therefore: −e− y = ex sin x + ex cos x − −e− y⎡⎣ ⎤⎦ + c

−e− y = ex sin x + ex cos x + e− y + c −2e− y = ex sin x + ex cos x + c

e− y = −12ex sin x + ex cos x⎡⎣ ⎤⎦ + c

Applying boundary conditions, let x = 0 and y = 0.

e− 0( ) = −12e 0( ) sin 0( ) + e 0( ) cos 0( )⎡⎣ ⎤⎦ + c

1 = −120 +1[ ] + c

⇒ c = 32

Therefore

e− y = −12ex sin x + ex cos x⎡⎣ ⎤⎦ +

32



C4 Vectors Problem 01 Question:

The line l1 has vector equation

r = 312

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ + λ

1−14

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

and the line l2 has vector equation

r = 04

−2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ + µ

1−10

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ ,

where λ and µ are parameters. The lines l1 and l2 intersect at the point B and the acute angle between l1 and l2 is θ. (a) Find the coordinates of B.

(4)

(b) Find the value of cos θ , giving your answer as a simplified fraction. (4)

The point A, which lies on l1, has position vector a = 3i + j + 2k. The point C, which lies on l2, has position vector c = 5i – j – 2k. The point D is such that ABCD is a parallelogram. (c) Show that ⎢ AB

⎢= ⎢ BC

⎢. (3)

(d) Find the position vector of the point D. (2)

Answer: a) At point B the vector equations for the lines are equal. Thus:

(This leads us to the equations)

[1]

[2] [3]

Please turn over…


Generally, one solves equations [1] & [2] simultaneously and substitute into equation [3] to find the values of λ and µ. Fortunately, this question has an easy route. Taking equation [3]:

Substituting this into the line equation for l1.

b) To find the angle, we use the equation:

It is important to note that the vectors that are dotted together must be the direction components of the lines only.

Let a equal the direction component of l1,

Let b equal the direction component of l2,

Please turn over...


Thus

c) (Substituting the vectors for a and b)

(Substituting the vectors for c and b)

Since both and equal , .

Please turn over…


d) This question is remarkably easy to answer and the technique used to answer this question arises often. But as always, it is a good idea to draw a diagram.

From the diagram above, it is easy to see that the position vector of D or if one prefers is simply:

Thus,


l1

.

O

C D

B A

l2

. .

.


C4 Vectors Problem 02

Question: Referred to a fixed origin O, the points A and B have position vectors (i + 2j – 3k) and (5i – 3j) respectively. (a) Find, in vector form, an equation of the line l1 which passes through A and B.

(2)

The line l2 has equation r = (4i – 4j + 3k) + µ (i – 2j + 2k), where µ is a scalar parameter. (b) Show that A lies on l2.

(1)

(c) Find, in degrees, the acute angle between the lines l1 and l2. (4)

The point C with position vector (2i – k) lies on l2. (d) Find the shortest distance from C to the line l1.

(4)

Answer: a) The line is given by the equation

Thus:

a) One can write line l2 as:

If point A lies on l2 then:

Please turn over…


Be equating coefficients we arrive at three equations: 1) 2) 3) Since for all three coefficients, point A exists on the line l2.

b) Using the formula

Using only the direction components of the two lines:

˚

Continued overleaf...


c) It is often best to draw a diagram to visualise how best to answer this question.

There are many ways in which one can answer this question, but the best way to answer this question is to use trigonometry. Since the shortest line linking point C to the line l1 will meet the line l1 at a right angle, we have a right angle triangle. Thus: Distance = (Thus determining )

units

Distance = Distance = 1.00 units


l2

O

A

C

l1

. . . Distance

19.5 ˚

Core 4 Workshop Download - Elite · PDF fileELITE Tuition, 1 Castle Road, Northolt,...

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